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Word problems: mix of 20% and 30% acids
- Thread starter firewolf
- Start date
>A chemist has one solution containing 20% acid, and a second containing 30% acid. How many liters of each solution must be combined to obtain 80 liters of a solution that is 28% acid?
Picture 3 containers. The first container has x liters of the 20% solution. The second container has y liters of the 30% solution. The third container is empty.
You are going to pour x liters of the 1st container and y liters of the 2nd container into the 3rd container so that you will have 80 liters of 28% acid. But we know that x+y=80. Therefore, we know that y=80-x
The key is to build equations that equate the amount of acid.
In other words, the amount of acid in the first container (that's .2x) plus the amount of acid in the second container (that's .3(80-x)) equals the amount of acid in the 3rd container (that's .28*80).
That should get you started.
Picture 3 containers. The first container has x liters of the 20% solution. The second container has y liters of the 30% solution. The third container is empty.
You are going to pour x liters of the 1st container and y liters of the 2nd container into the 3rd container so that you will have 80 liters of 28% acid. But we know that x+y=80. Therefore, we know that y=80-x
The key is to build equations that equate the amount of acid.
In other words, the amount of acid in the first container (that's .2x) plus the amount of acid in the second container (that's .3(80-x)) equals the amount of acid in the 3rd container (that's .28*80).
That should get you started.